Note: Question was originally to solve it algebraically, though I've decided to change it to analytically due to the comments and answers.
When trying to solve $\sin(x)=x$, the obvious first solution is $x=0$. There are, however, an infinite amount of complex values of $x$ we can try to find. However, we are going to ignore these.
I was wondering if there was a way to analytically solve for $x$ in $\sin(x)=x$. It does not appear to be possible, just like we can't solve $\cos(x)=x$ analytically or easily, but since $\sin(x)=x$ has such a simple exact answer, I wondered if there is a way you could do it.
So does there exist an analytic way we can solve this? If so, how? If not, how else would we solve it other than graphically?
Best Answer
If the problem could be solved by purely algebraic means (with a finite number of steps), that would imply that $\sin(x)$ could be given a polynomial representation from which you could go about your usual routine of factoring to find the zeroes of the polynomial.
The interesting point here is that no such representation for $\sin(x)$ exists, unless you are okay with it being infinitely long.
The trigonometric functions like $\sin()$ and $\cos()$ are part of a category of transcendental functions--so called because they transcend the expressive power of algebra to describe them.
Here's a shot at solving it algebraically if we can cheat and use a result from calculus:
Given this identity:$$\sin(x) = x - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac{x^7}{7!} + \cdots $$
Subtract out your problem $\sin(x) = x$
$$0 = - \frac {x^3}{3!} + \frac {x^5}{5!} - \frac{x^7}{7!} + \cdots $$
$$0 = x^3(- \frac {1}{3!} + \frac {x^2}{5!} - \frac{x^4}{7!} + \cdots) $$
$$x^3 = 0 \quad \mathrm{or} \quad (- \frac {1}{3!} + \frac {x^2}{5!} - \frac{x^4}{7!} + \cdots) = 0 $$
So now we have our "algebraic solution" that $x = 0$.